Question

A pitcher throws a baseball at 31.0 m/s. The batter hits it and gives it a velocity of 40.0 m/s in the opposite direction. If the ball and bat are in contact for 4.00 x 10^-2 s, what is the average acceleration of the ball during this interval?

Answer 1

Acceleration will be 1775
`m/sec^2`

Step-by-step explanation:

We have given that pitcher throws the ball with speed of 31 m/sec

And the ball bounce back with a speed of 40 m/sec

Time
`t=4* 10^(-2)sec`

From first equation of motion we know that v = u+at

So acceleration
`a=(v-u)/(t)=(40-(-31))/(4* 10^(-2))=1775m/sec^2` (for calculation we take -31 because it is in opposite direction )